Some mid-sized ¿hyperbolic? manifolds and SnapPea - MathOverflow

Some mid-sized ¿hyperbolic? manifolds and SnapPea

2012-11-14T23:47:19Z

I've recently come across some mid-sized 3-manifolds that I think are likely hyperbolic, but SnapPea has some trouble with them. This is related to my previous question

http://mathoverflow.net/questions/4918/can-you-fool-snappea

but in this case I'm dealing with closed, orientable 3-manifolds instead of knot and link complements.

The 3-manifolds I've come across have 11, 12 and 13 tetrahedra in their triangulations. One of them SnapPea finds a solution to the gluing equations but it has "negatively oriented tetrahedra". Does this mean what I think it means -- that once you've put the geometric structure on the tetrahedra, you have a tetrahedron folded-over? If you view the gluing equations from the upper half-space model, they say that the sum of a bunch of angles should be $2\pi$. Is this the case where one of those angles is negative?

The other two triangulations SnapPea finds geometric structures with degenerate tetrahedra. In the upper half-space model this is where one of the angles is zero, I believe. Is there a way to fix this, so that I could get a Dirichlet domain, drill and fill, etc? This is my primary question.

Here are the triangulations, both in SnapPea format and Regina format.

Regina file, all triangulations

11-tet, SnapPea

12-tet, SnapPea

13-tet, SnapPea

With the 12-tetrahedron example, SnapPea can generate a Dirichlet Domain. This one has what looks almost like bevelled-edges. Is there a way to find a better basepoint to grow the Dirichlet domain from?

edit: After working a bit with Nathan's answer, I've identified these three manifolds and got a little closer to understanding how to work SnapPea to maximal advantage.

tri11, as Nathan mentioned, is hyperbolic. It has a fairly pretty Dirichlet domain.

Another common name for this manifold would be the 0-surgery on the 2-component link $7a_6$ (in the Thistlethwaite table). Similarly tri12 can be identified as Nathan says.

After playing around with Regina a bit I found an incompressible torus in tri13 that splits tri13 into the union of an orientable $I$-bundle over the Klein bottle and a figure-8 complement. So this answers the core of my question.

Answer by Nathan Dunfield for Some mid-sized ¿hyperbolic? manifolds and SnapPea

2012-11-16T01:10:33Z

The first two of these manifolds are hyperbolic. The problem, I suspect, is that they are being specified as Dehn fillings on 1-cusped manifolds with essential tori. So one needs to pick a different knot in the closed manifold to actually see the hyperbolic structure. One way to force SnapPy to do this is to create a 1-vertex non-ideal triangulation and then use one of the edges there as the knot. For example:

sage: M = Manifold('tri11.txt')
sage: N = Manifold(M.filled_triangulation()._to_string())
sage: N.solution_type()
'all tetrahedra positively oriented'
sage: closed = snappy.OrientableClosedCensus()
sage: closed.identify(N)
m038(1,2)

The tri12 manifold is "m032(5,2)". I strongly suspect the tri13 manifold is not hyperbolic. Instead, I suspect it has a nontrivial JSJ splitting where one piece is the figure-8 complement, since sometimes the volume appears to be 2.0988...

Answer by Bruno Martelli for Some mid-sized ¿hyperbolic? manifolds and SnapPea

2012-11-16T10:06:40Z

That's just a long comment about tri13. I don't have Regina now, but you can get some informations by looking only at the 1-skeleton $G$ of the closed triangulation for tri13. Among the (experimentally) minimal triangulations for tri13, i.e. the ones that you find by trying to simplify it as much as you (or Regina) can, you probably find one where the 1-skeleton $G$ contains a pair of disconnecting edges, which split the $G$ into two parts $G_1$ and $G_2$. In any experimentally minimal triangulation such a pair of edges is always transverse to a torus $T$, which cut the manifold $M$ into two parts $M_1$ and $M_2$, containing $G_1$ and $G_2$ correspondingly.

According to Nathan's suggestion, the torus $T$ should separate a figure-8 knot complement (or its sibling) $M_1$ and a small Seifert space $M_2$. You can see this from the 1-skeleton $G$: the portion $G_1$ must have at least 8 vertices to be hyperbolic, and it has probably 8 or 9 (corresponding to the figure-8 knot sibling and the figure-8 knot complement), and the other portion $G_2$ has of course 5 or 4 vertices. If $G_2$ is layered, then the torus is compressible and you have a Dehn fillig of $M_1$. If (as we expect) $G_2$ is not layered, then it is certainly not a solid torus (minimal triangulations of solid tori with small number of tetrahedra must be layered, as far as I remember).

Summing up, I think that:

if a triangulation for $M$ is experimentally minimal with a reasonable number of tetrahedra and its 1-skeleton $G$ decomposes into two pieces $G_1$ and $G_2$, noone of which is layered, then you can conclude theoretically that it is not hyperbolic.

The numer of tetrahedra must be reasonable, because minimal triangulations of solid tori are classified only for small number of tetrahedra (I don't know how many...)

A fool-proof alternative is to use Matveev's Regina, which tells you exactly the JSJ decomposition of the manifold.